Please give me a hint to solving this simple vector dot product proof

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Homework Statement



Let 'u' and 'v' be two non zero vectors such that the prjection of 'u' along 'v' equals the projection of 'v' along 'u.' Using the formula for projection, show that 'u' and 'v' are either perpendicular or parallel.

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The Attempt at a Solution



Please don't just answer it, I would like to do this one on my own. But I first need a hint because I have been trying for about 30 minutes.

I know that the projection of 'u' along 'v' is u dot v, divided by the square of the norm of 'v'. Then this scalar is multiplied through 'v'. But that's about all I have.

Edit: I guess I said more about how far I got. I get the following:

[itex]\frac{1}{||v||^{2}}[/itex][itex]\overline{v}[/itex]=[itex]\frac{1}{||u||^{2}}[/itex][itex]\overline{u}[/itex]
 
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  • #2
LCKurtz
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Homework Statement



Let 'u' and 'v' be two non zero vectors such that the prjection of 'u' along 'v' equals the projection of 'v' along 'u.' Using the formula for projection, show that 'u' and 'v' are either perpendicular or parallel.

Please don't just answer it, I would like to do this one on my own. But I first need a hint because I have been trying for about 30 minutes.

I know that the projection of 'u' along 'v' is u dot v, divided by the square of the norm of 'v'. Then this scalar is multiplied through 'v'. But that's about all I have.

Edit: I guess I said more about how far I got. I get the following:

[itex]\frac{1}{||v||^{2}}[/itex][itex]\overline{v}[/itex]=[itex]\frac{1}{||u||^{2}}[/itex][itex]\overline{u}[/itex]
Think about what the identity $$
\vec A\cdot \vec B = |\vec A||\vec B|\cos\theta$$implies about this problem.
 
  • #3
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(sorry for the long time before replying)

Thanks! That hint definitely helped me out!

In case anyone comes by this thread seeking the same as I did, when you apply the identity that LCKurtz showed us, you get a scalar number multiplied by vector v equals vector u. Obviously, this means they are either parallel or anti-parallel.
 
  • #4
LCKurtz
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(sorry for the long time before replying)

Thanks! That hint definitely helped me out!

In case anyone comes by this thread seeking the same as I did, when you apply the identity that LCKurtz showed us, you get a scalar number multiplied by vector v equals vector u. Obviously, this means they are either parallel or anti-parallel.
Unless ##\cos\theta=0##, in which case ...
 
  • #5
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But then wouldn't that just mean that they are parallel anyways? So no matter how you look at it, if proj u over v equals proj v over u, as long as these aren't zero vectors, wouldn't the angle HAVE to be 0 or 180?
 
  • #6
LCKurtz
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Homework Statement



Let 'u' and 'v' be two non zero vectors such that the prjection of 'u' along 'v' equals the projection of 'v' along 'u.' Using the formula for projection, show that 'u' and 'v' are either perpendicular or parallel.
Unless ##\cos\theta=0##, in which case ...
But then wouldn't that just mean that they are parallel anyways? So no matter how you look at it, if proj u over v equals proj v over u, as long as these aren't zero vectors, wouldn't the angle HAVE to be 0 or 180?
What do you get for ##\vec i \cdot \vec j##?
 
  • #7
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What do you get for ##\vec i \cdot \vec j##?
Oh I see.. so then given no extra information about u and v, they both can be EITHER parallel OR perpendicular? I guess I misread the question.. I thought it was asking to show that it was one OR the other, and because of this, I blindly looked for one until I found it, and then stopped before I checked the other.

Definitely one vector dotted into another equals 0 if they are both perpendicular.

Thanks again!
 

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